The given figures shows a triangle ABC in which AD is perpendicular to side BC and BD = CD. Prove that : (i) $$ \Delta ABD \cong \Delta ACD $$ (ii) $$ AB = AC $$ (iii) $$ \angle B = \angle C $$
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Q2
Prove that $$ \Delta ABD \cong \Delta ACD $$
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Q3
ABC is an isosceles triangle with AB =AC and D is a point on BC such that AD⊥BC (Fig. 7.13). To prove that ∠BAD=∠CAD, a student proceeded as follows: ΔABD and ΔACD, AB = AC (Given) ∠B=∠C (because AB = AC) and ∠ADB=∠ADC Therefore, ΔABD≅ΔACD(AAS) So, ∠BAD=∠CAD(CPCT) What is the defect in the above arguments?
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Q4
In the given figure , prove that $$ \Delta ABD \cong \Delta ACD $$
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Q5
Using distance formula prove that the following points are collinear:
(i) A(4, –3, –1), B(5, –7, 6) and C(3, 1, –8)
(ii) P(0, 7, –7), Q(1, 4, –5) and R(–1, 10, –9)
(iii) A(3, –5, 1), B(–1, 0, 8) and C(7, –10, –6)