Question

$ABCD$ is a square. $E$ and $F$ are points respectively on $BC$ and $CD$ such that $\angle EAF = 45^\circ$.

score 1.6k · Answer 1 · 2020-10-16 09:56:12

1.6k

Views

$ABCD$ is a square. $E$ and $F$ are points respectively on $BC$ and $CD$ such that $\angle EAF = 45^\circ$.

Published on 16 Oct 2020 - 9:56

score 424 · Answer 2 · 2020-10-16 15:00:00

424

Views

$ABCD$ is a square. $E$ is the midpoint of $CB$, $AF$ is drawn perpendicular to $DE$. If the side of the square is $2016$ cm , find $BF$.

Published on 16 Oct 2020 - 15:00

score 118 · Answer 3 · 2020-10-16 17:23:00

118

Views

$ABCD$ is a rectangle of area $210$ cm$^2$. $L$ is a mid-point of $CD$ . $P,Q$ trisect $AB$ . $AC$ cuts $LP,LQ$ at $M,N$ respectively.

Published on 16 Oct 2020 - 17:23

score 722 · Answer 4 · 2020-10-22 14:01:33

722

Views

In an isosceles right $\Delta ABC$, $\angle B = 90^\circ$. AD is the median on BC. Let $AB = BC = a$.

Published on 22 Oct 2020 - 14:01

score 449 · Answer 5 · 2020-10-25 14:49:33

449

Views

In right $\Delta ABC$, $\angle C = 90^\circ$. $E$ is on $BC$ such that $AC = BE$. $D$ is on $AB$ such that $DE \perp BC$ .

Published on 25 Oct 2020 - 14:49

score 160 · Answer 6 · 2020-10-26 11:48:45

160

Views

In $\Delta ABC$, $AC = BC$ and $\angle C = 120^\circ$. $M$ is on side $AC$ and $N$ is on side $BC$ .

Published on 26 Oct 2020 - 11:48

score 373 · Answer 7 · 2020-10-26 12:34:44

373

Views

In $\Delta ABC$, angle bisector of $\angle ABC$ and median on side $BC$ intersect perpendicularly

Published on 26 Oct 2020 - 12:34

score 978 · Answer 8 · 2020-10-26 16:54:14

978

Views

When extending the sides $AB,BC,CA$ of $\Delta ABC$ to $B',C',A'$ respectively, such that $AB' = 2AB$ , $CC' = 2BC$ , $AA' = 3CA$ .

Published on 26 Oct 2020 - 16:54

score 84 · Answer 9 · 2020-10-30 23:22:49

84

Views

How to prove that a triangle is uniquely determined by an angle, its opposite side and its perpendicular height.

Published on 30 Oct 2020 - 23:22

score 778 · Answer 10 · 2020-10-31 11:18:00

778

Views

In $\Delta ABC$, $AB:AC = 4:3$ and $M$ is the midpoint of $BC$ . $E$ is a point on $AB$ and $F$ is a point on $AC$ such that $AE:AF = 2:1$

Published on 31 Oct 2020 - 11:18

score 680 · Answer 11 · 2020-11-01 11:18:55

680

Views

In $\triangle ABC, AB = 28, BC = 21$ and $CA = 14$. Points $D$ and $E$ are on $AB$ with $AD = 7$ and $\angle ACD = \angle BCE$

Published on 01 Nov 2020 - 11:18

score 451 · Answer 12 · 2020-11-08 11:50:44

451

Views

In $\Delta ABC$, $AB = 14, BC = 16, AC = 26$. $M$ is the midpoint of $BC$ and $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$.

Published on 08 Nov 2020 - 11:50

score 25 · Answer 13 · 2020-11-09 21:46:14

25

Views

Planar convex sets whose self-intersections are similar to themselves

Published on 09 Nov 2020 - 21:46

score 594 · Answer 14 · 2020-11-11 15:34:24

594

Views

Let $O$ be the centre of the circumcircle of $\Delta ABC$, $P$ and $Q$ be the midpoint of $AO$ and $BC$, respectively.

Published on 11 Nov 2020 - 15:34

score 394 · Answer 15 · 2020-11-13 17:21:46

394

Views

In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$.

Published on 13 Nov 2020 - 17:21

List Question

$ABCD$ is a square. $E$ and $F$ are points respectively on $BC$ and $CD$ such that $\angle EAF = 45^\circ$.

$ABCD$ is a square. $E$ is the midpoint of $CB$, $AF$ is drawn perpendicular to $DE$. If the side of the square is $2016$ cm , find $BF$.

$ABCD$ is a rectangle of area $210$ cm$^2$. $L$ is a mid-point of $CD$ . $P,Q$ trisect $AB$ . $AC$ cuts $LP,LQ$ at $M,N$ respectively.

In an isosceles right $\Delta ABC$, $\angle B = 90^\circ$. AD is the median on BC. Let $AB = BC = a$.

In right $\Delta ABC$, $\angle C = 90^\circ$. $E$ is on $BC$ such that $AC = BE$. $D$ is on $AB$ such that $DE \perp BC$ .

In $\Delta ABC$, $AC = BC$ and $\angle C = 120^\circ$. $M$ is on side $AC$ and $N$ is on side $BC$ .

In $\Delta ABC$, angle bisector of $\angle ABC$ and median on side $BC$ intersect perpendicularly

When extending the sides $AB,BC,CA$ of $\Delta ABC$ to $B',C',A'$ respectively, such that $AB' = 2AB$ , $CC' = 2BC$ , $AA' = 3CA$ .

How to prove that a triangle is uniquely determined by an angle, its opposite side and its perpendicular height.

In $\Delta ABC$, $AB:AC = 4:3$ and $M$ is the midpoint of $BC$ . $E$ is a point on $AB$ and $F$ is a point on $AC$ such that $AE:AF = 2:1$

In $\triangle ABC, AB = 28, BC = 21$ and $CA = 14$. Points $D$ and $E$ are on $AB$ with $AD = 7$ and $\angle ACD = \angle BCE$

In $\Delta ABC$, $AB = 14, BC = 16, AC = 26$. $M$ is the midpoint of $BC$ and $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$.

Planar convex sets whose self-intersections are similar to themselves

Let $O$ be the centre of the circumcircle of $\Delta ABC$, $P$ and $Q$ be the midpoint of $AO$ and $BC$, respectively.

In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$.

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