In the given figure- - 1 - - 2 and AC-BD - CB-CE- Prove that - ACB - DCE-

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In the given ...

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In the given figure, $∠ 1 = ∠ 2$ and $\frac{A C}{B D} = \frac{C B}{C E}$ .

Prove that $Δ A C B \sim Δ D C E$ .

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∠ 1 = ∠ 2 and \frac{A C}{B D} = \frac{C B}{C E} .

∠ 1 = ∠ 2

\frac{A C}{B D} = \frac{C B}{C E}

Δ

\sim Δ

In Fig. BD and CE are two altitudes of a $Δ$ ABC such that BD = CE.

Prove that the ABC is an isoceles triangle. [2 MARKS]

∠ A C B = 90^{0}

C D ⊥ A B

\frac{{C B}^{2}}{{C A}^{2}} = \frac{B D}{A D}

∠ B C A = 2 ∠ B A C

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