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Prove that : (i) \( \Delta \mathrm { ABD } \equiv \Delta \mathrm { ACD } \) (ii) \( \angle B = \angle C \) (iii) \( \angle \mathrm { ADB } = \angle \mathrm { ADC } \) \( \frac { 1 } { 8 } + \frac { 1 } { 0 } + \frac { 1 } { 0 } \)

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Similar Questions
Q1
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 ADBC (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, ΔABDΔACD(AAS)
So, BAD=CAD(CPCT)
What is the defect in the above arguments?
78853_330861415ee64345b8ba219aaf8ae2ec.png
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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)
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