Question

In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC and AC . Point G ∈ AB so that DG ⊥ AB and H ∈ AB so that EH ⊥ AB . Prove that ΔCEH and ΔCDG are isosceles.

Answer 1

Proof: $\triangle CEH$ and $\triangle CDG$ are isosceles triangles.

Explanation: Given in $\triangle ABC$ D and E are angle bisector of   $\angle A$and $\angle B$ respectively.

Where G and H are points in AB such that $DG\perp AB$ and $EH\perp AB$.

Let us take two triangles $\triangle BCE$ and $\triangle BEH$

$\angle BCE=\angle BHE$ (Right angles)

BE=BE, (common segment)

$\angle HBE=\angle CBE$ ( Because BE is angle bisector)

Thus, $\triangle BCE\cong\triangle BEH$(ASA)

Therefore, EH= CE (CPCT)

So, in $\triangle CEH$, EH=CE ⇒$\triangle CEH$ is an isosceles triangle.

Now, in $\triangle ACD$ and $\triangle ADG$,

$\angle ACD=\angle AGD$ (Right angles)

AD=AD (common segment)

$\angle CAD=\angle DAG$ ( Because AD is angle bisector)

⇒$\triangle ADC\cong\triangle AGD$ (ASA)

Thus, CD=DG (CPCT)

So, in $\triangle CDG$, CD=DG ⇒$\triangle CDG$ is an isosceles triangle.