In the figure given below-Then prove that ACR - YCR-

Taking nbsp;ACR and YCR, we have AR = YR (perpendicular bisector) nbsp;[1 Mark] ∠ ARC = ∠ YRC = 90^∘ nbsp;(perpendicular bisector) nbsp;[1 Mark] CR ...

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Standard IX

Mathematics

Construction of Triangle 2

In the figure...

Question

In the figure given below,

BD is the perpendicular bisector of AX

CF is the perpendicular bisector of AY

Then prove that $△$ ACR $≅$ $△$ YCR.

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Solution

taking $△$ ACR and $△$ YCR, we have

AR = YR (perpendicular bisector) $[1 M a r k]$

$∠$ ARC = $∠$ YRC = $90^{\circ}$ (perpendicular bisector) $[1 M a r k]$

CR = CR (common side)

$△$ ACR $≅$ $△$ YCR (By SAS Congruency) $[1 M a r k]$

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In the figure given below, BD is the perpendicular bisector of AX CF is the perpendicular bisector of AY Then prove that △ ACR ≅ △ YCR.

taking △ ACR and △ YCR, we have AR = YR (perpendicular bisector) [1 Mark] ∠ ARC = ∠ YRC = 90∘ (perpendicular bisector) [1 Mark] CR = CR (common side) △ ACR ≅ △ YCR (By SAS Congruency) [1 Mark]

In the figure given below,

BD is the perpendicular bisector of AX

CF is the perpendicular bisector of AY

Then prove that $△$ ACR $≅$ $△$ YCR.

taking $△$ ACR and $△$ YCR, we have

AR = YR (perpendicular bisector) $[1 M a r k]$

$∠$ ARC = $∠$ YRC = $90^{\circ}$ (perpendicular bisector) $[1 M a r k]$

CR = CR (common side)

$△$ ACR $≅$ $△$ YCR (By SAS Congruency) $[1 M a r k]$