Respuesta :

msm555

Answer:

Solution given:

AB parallel BH parallel CH

1.

In triangle ∆AFD ,∆BGD ,∆CHD

<AFD=<BGD=<CHD

[being corresponding angle]

<DAF=<DBG=<DCH

[being corresponding angle]

<D is common for all.

So

∆AFD ~∆BGD ~∆CHD

[By A .A .A. axiom]

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