Hi everyone,
for the situation in the sketch above it is given that $|PC|\geq\ $|PA|=|PB| and $|PD| \geq \ $|PB|. I need to prove that $|CD| \geq \ $|AB|.
Without loss of generality I am assuming that $|PD| \geq \ $|PC|. It feels like I need to use intercept theorem and/or triangle inequality but I am getting nowhere.
Can anyone help, please?
Thanks a lot!
On
Put everything in coordinate system. Say $P(0,0)$.
Suppose $A,C$ are on line $y=-kx$ and $B,D$ are on line $y=kx$. Then $A(a,-ka)$ and $B(a,ka)$ and $C(c,-kc)$ and $D(d,kd)$. WLOG we can assume $a=1$ and $c,d>1$.
Then $AB = 2k$ and $$ CD = \sqrt{(c-d)^2+(c+d)^2k^2}$$
We have to check whether $$(c-d)^2+(c+d)^2k^2> 4k^2$$
Since $c+d>2$ we see that this is true.
On
Assume WLOG $PD\ge PC$ and draw point $E$ on $BD$ such that $PE=PC$. Triangles $PAB$ and $PCE$ are similar, and from $CP>AP$ it follows $CE>AB$.
Consider then triangle $CDE$: angle $\angle CED$ is obtuse, because its adjacent $\angle PEC$ is acute (base angle of isosceles triangle $PCE$). Hence $CD>CE$, as in a triangle greater side is opposite to greater angle.
On
Assume wlog that $PD>PC$. Let $C'$ lie on $BD$ so that $CC'$ is parallel to $AB$
Then triangles $ABP$ and $CC'P$ are similar, and $$\frac{CC'}{AB}=\frac{PC}{PA}>1\implies CC'>AB$$
Furthermore, angle $CC'D=90+\frac 12$angle $APB$ and is therefore obtuse, so that $\cos CC'D$ is negative.
Therefore by the Cosine Rule, $DC^2>CC'^2+C'D^2$
Thus $DC>CC'>AB$
Let $K\in BD$ such that $AK||DC$.
Thus, since $\measuredangle ABK>90^{\circ},$ we obtain: $$DC>AK>AB$$ and we are done!