In the given figure- - 1- 2 and ACBD - CBCE- Prove that A C B - D C E-

We have: ACBD nbsp;= nbsp;CBCE⇒ACCB nbsp;= nbsp;BDCE⇒ACCB nbsp;= nbsp;CDCE nbsp; nbsp;(Since, nbsp;BD = nbsp;DC nbsp;as nbsp;∠1 nbsp;= ...

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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 ∆ ACB ∼ ∆ DCE.

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

Prove that $Δ A C B \sim Δ D C E$ .

∠ 1 = ∠ 2

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

Δ

\sim Δ

A C ⊥ B D

C E ⊥ A D

A C^{2} = D A \times A E

In the given figure, $O$ is the midpoint of each of the line segments $A B$ and $C D .$ Prove that $A C = B D$ and $A C ∥ B D .$

∠ A C B = 90^{0}

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

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