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.

Step-by-step 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.