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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parmesanchilliwack parmesanchilliwack · Answer 1 · 2018-11-19T07:36:31+01:00

Proof: [tex]\triangle CEH[/tex] and [tex]\triangle CDG[/tex] are isosceles triangles.

Explanation: Given in [tex]\triangle ABC[/tex] D and E are angle bisector of [tex]\angle A[/tex]and [tex]\angle B[/tex] respectively.

Where G and H are points in AB such that [tex]DG\perp AB[/tex] and [tex]EH\perp AB[/tex].

Let us take two triangles [tex]\triangle BCE[/tex] and [tex]\triangle BEH[/tex]

[tex]\angle BCE=\angle BHE[/tex] (Right angles)

BE=BE, (common segment)

[tex]\angle HBE=\angle CBE[/tex] ( Because BE is angle bisector)

Thus, [tex]\triangle BCE\cong\triangle BEH[/tex](ASA)

Therefore, EH= CE (CPCT)

So, in [tex]\triangle CEH[/tex], EH=CE ⇒[tex]\triangle CEH[/tex] is an isosceles triangle.

Now, in [tex]\triangle ACD[/tex] and [tex]\triangle ADG[/tex],

[tex]\angle ACD=\angle AGD[/tex] (Right angles)

AD=AD (common segment)

[tex]\angle CAD=\angle DAG[/tex] ( Because AD is angle bisector)

⇒[tex]\triangle ADC\cong\triangle AGD[/tex] (ASA)

Thus, CD=DG (CPCT)

So, in [tex]\triangle CDG[/tex], CD=DG ⇒[tex]\triangle CDG[/tex] is an isosceles triangle.