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Congruence — MATHS STD 7 — Question

Gujarat BoardEnglish MediumSTD 7MATHSCongruence3 Marks
Question
In the adjoining figure, $\text{AB}=\text{AC}$ and $\text{BD}=\text{DC}.$ Prove that $\triangle\text{ADB}\cong\triangle\text{ADC}$ and hence show that.
$i. \angle\text{ADB}=\angle\text{ADC}=90^\circ$
$ii. \angle\text{BAD}=\angle\text{CAD}$
✓

Answer

In $\triangle\text{ABD}$ and $\triangle\text{ACD},$
$\text{AD}=\text{AD} ($common$)$
$\text{AB}=\text{AC} ($given$)$
$\triangle\text{ABD}\cong{}\triangle\text{ADC} (\ce{SSS}$ condition$)$
$\angle\text{BAD}=\angle\text{CAD} (c.p.c.t.)$ and
$\angle\text{ADB}=\angle\text{ADC} (c.p.c.t.)$ But
$\angle\text{ADB}+\angle\text{ADC}=180^\circ ($Linear pair$)$
$\angle\text{ADB}=\angle\text{ADC}=90^\circ$ Hence proved.

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