Question-216958

Question Number 216958 by abdi last updated on 25/Feb/25

Answered by A5T last updated on 25/Feb/25

Commented by A5T last updated on 26/Feb/25

2)

Let circle passing through CHB and ACB be

Ω and Γ respectively

∠ CHB = 90 ° \Rightarrow CB is the diameter of Ω

∠ ACB = 90 ° \Rightarrow AC is tangent to Ω

Power of a point with respect to Ω from A

\Rightarrow {AC}^{2} = AH \times AB .

∠ ACB = 90 ° \Rightarrow AB is the diameter of Γ

∠ CHB = 90 ° \Rightarrow AB bisects CD \Rightarrow CH = HD

Power of a point with respect to Γ from H

\Rightarrow AH \times BH = CH \times HD = {CH}^{2}

Answered by A5T last updated on 25/Feb/25

Commented by A5T last updated on 25/Feb/25

3)

Van {Aubel}^{'} s theorem : \frac{AG}{GF} = \frac{AD}{DB} + \frac{AE}{EC}

\Rightarrow \frac{AG}{GF} = \frac{1}{1} + \frac{1}{1} = 2 \Rightarrow AG = 2 GF

Similarly, we get BG = 2 GE; CG = 2 DG

Answered by A5T last updated on 25/Feb/25

Commented by A5T last updated on 25/Feb/25

{Menelaus}^{'} Theorem : \frac{{AC}_{1}}{C_{1} B} \times \frac{BC}{{CA}_{1}} \times \frac{A_{1} D}{DA} = 1

\Rightarrow \frac{1}{1} \times \frac{3}{1} \times \frac{A_{1} D}{DA} = 1 \Rightarrow AD : {DA}_{1} = 3 : 1

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