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ahrayia [7]
4 years ago
12

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 an

d AC . Point G ∈ AB so that DG ⊥ AB and H ∈ AB so that EH ⊥ AB . Prove that ΔCEH and ΔCDG are isosceles.

Mathematics
1 answer:
alexira [117]4 years ago
4 0

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

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.


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