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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.

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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 Aand
\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.


In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of-example-1
User Biggy Smith
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