Geometry :Triangles

Question

In the figure above, $AB = 18$, $AC = 24$, $BC = 30$. $BF$ and $CF$ bisect angles $ABC$ and $ACB$ respectively and $DH\parallel BC$. What is the perimeter of triangle $ADH$?

The answer is $42$. But I’m not sure how ?

Answer 1

HINTS:

• $DH||BC$ means that you have similar triangles. Let $r=AD/AB$
• From the point above, $\angle ADH=\angle ABC$
• What do you know about exterior angles? Compare $\angle ABF$ and $\angle DFB$

The above points will let you calculate $r$. Can you take it from here? If not, describe in details what you have tried, based on these hints.

Extra: Since $\angle ADH=\angle ABC$ and $\angle ADH$ is exterior to $\triangle BDF$, it means $\angle ADH=\angle DBF+\angle DFB$. But $\angle DBF=\frac12\angle ABC$, so $\angle DFB=\angle ABC-\frac12\angle ABC$. So $\triangle BDF$ is isosceles. Therefore $DB=DF$. Similarly $FH=HC$.

Next, use $r=AD/AB=AH/HC$. Write $BD$ in terms of $AB$ and $r$. Do the similar thing for $HC$. Now you can write $DH$ in terms of $AB$, $AC$, and $r$. Also, $r=DH/BC$. So you can find $r$.

Alternative solution:

Once you know $DB=DF$ and $FH=HC$, then the perimeter of $\triangle ADH$ is : $$AD+DH+HA=AD+DF+FH+HA=AD+DB+CH+HA=AB+CA=42$$

Answer 2

Semi-perimeter of $\triangle ABC$ is $s=\frac{a+b+c}{2}=\frac{30+24+18}2=36$.

By Heron's formula the area of $\triangle ABC$ is $(ABC)=\sqrt{s(s-a)(s-b)(s-c)}=6^3.$

The inradius of $\triangle ABC$ is $r=\frac{(ABC)}{s}=6.$

The altitude of side $a=BC$ is $h_a=\frac{2(ABC)}a=\frac{72}5.$

The similarity ratio of $\triangle ADH$ to $\triangle ABC$ is $k=\frac{h_a-r}{h_a}=\frac7{12}.$

The perimeter of $\triangle ADH$ is $k$ times the perimeter of $\triangle ABC$: $\frac7{12}\times72=42.$

Answer 3

Since the sides have the Pythagorian triple ratio of 3:4:5 the triangle is right angled triangle.

A right angle triangle has area equal to half the product of the two sides adjacent to the hypotenuse so area = (AB.AC)/2. area = 216.

For any triangle, the semi perimeter (s) = perimeter/2 = (AB+AC+BC)/2. s = 36.

For any triangle, the inradius (r), which is the perpendicular distance from any side to where the angle bisectors intersect is given by area/s. r = 6.

This is the height of the segment DH above the parallel segment BC.

For a right angled triangle, the height (h) from the hypotenuse to the right angle is 2 area/hypotenuse = 432/BC. h = 72/5.

Since the triangles are similar:

$\frac{(h-r)}{h} = \frac{AD}{AB} = \frac{AH}{AC} = \frac{DH}{BC} = \frac 7 {12}$.

Since all the sides are scaled by the same proportion the perimeter of the smaller right angled triangle is $\frac 7 {12} (AB+AC+BC) = 42$.