(C) In $\Delta ABC$ and $\Delta ADB$:$1$. $m \angle ABC = m \angle ADB = 90^{\circ}$ (Given).$2$. $m \angle BAC = m \angle DAB$ (Common angle).By the

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(C) In $\Delta ABC$ and $\Delta ADB$:$1$. $m \angle ABC = m \angle ADB = 90^{\circ}$ (Given).$2$. $m \angle BAC = m \angle DAB$ (Common angle).By the $AA$ (Angle-Angle) similarity criterion,the triangles are similar.Therefore,the correspondence $ABC \leftrightarrow ADB$ is a similarity.

Class 10 Mathematics Triangles Mix Examples - Triangles

Medium

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. Then,the correspondence $ABC \leftrightarrow \ldots$ between $\Delta ABC$ and $\Delta ADB$ is a similarity.

A
$ABD$
B
$BDA$
C
$ADB$
D
$BAD$

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Triangles

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Mix Examples - Triangles

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $BM = 2 CM$,prove that $AC = 5 CM$.

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is a median. If $AB = 15$ and $BC = 20$,find $BM$.

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In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$ and $\overline{AD}$ is a median. If $BC = 12$ and $AD = 8$,find $AB$.

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In $\square ABCD$,$A-Q-B$,$A-P-C$ and $A-R-D$. Given $\overline{PQ} \parallel \overline{BC}$ and $\overline{PR} \parallel \overline{DC}$. Prove that $\frac{AR}{AD} = \frac{AQ}{AB}$.

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In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 3x - 1$,$MB = 2x + 1$,$AN = 3x + 1$ and $NC = 5x + 1$,find the value of $x$.

Medium

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. Then,the correspondence $ABC \leftrightarrow \ldots$ between $\Delta ABC$ and $\Delta ADB$ is a similarity.

Similar Questions

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $BM = 2 CM$,prove that $AC = 5 CM$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is a median. If $AB = 15$ and $BC = 20$,find $BM$.

In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$ and $\overline{AD}$ is a median. If $BC = 12$ and $AD = 8$,find $AB$.

In $\square ABCD$,$A-Q-B$,$A-P-C$ and $A-R-D$. Given $\overline{PQ} \parallel \overline{BC}$ and $\overline{PR} \parallel \overline{DC}$. Prove that $\frac{AR}{AD} = \frac{AQ}{AB}$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 3x - 1$,$MB = 2x + 1$,$AN = 3x + 1$ and $NC = 5x + 1$,find the value of $x$.

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