In ABC, if AD is perpendicular to BC and
AD2 = BD X DC then
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Answer:
10th
Maths
Triangles
Criteria for Triangle Similarity
In triangle ABC, AD is perp...
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Asked on December 30, 2019 by
Noor Punjabi
In triangle ABC, AD is perpendicular to BC and AD
2
=BD×DC. Find ∠BAC
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ANSWER
Given: △ABC, AD
2
=BD×DC, AD⊥BC
Using Pythagoras Theorem
In Δ ABD
AB
2
=AD
2
+BD
2
AB
2
=BD×DC+BD
2
[GivenAD
2
=BD×DC]
AB
2
=BD×(DC+BD)
AB
2
=BD×BC .......... (i)
In Δ ADC,
AC
2
=AD
2
+DC
2
AC
2
=BD×DC+DC
2
=DC×(BD+DC)
AC
2
=DC×BC ..........(ii)
∴AB
2
+AC
2
=BD×BC+DC×BC [Adding (i) and (ii)]
AB
2
+AC
2
=BC×(BD+DC)
AB
2
+AC
2
=BC×BC=BC
2
⇒∠BAC=90
∘
[Using converse of Pythagoras Theorem]
Answered by
0
Step-by-step explanation:
Given : In triangle ABC , AD is perpendicular to BC and AD² = BDxDC
To prove : BAC = 90°
Proof : in right triangles ∆ADB and ∆ADC
So, Pythagoras theorem should be apply ,
Then we have ,
AB² = AD² + BD² ----------(1)
AC²= AD²+ DC² ---------(2)
AB² + AC² = 2AD² + BD²+ DC²
= 2BD . CD + BD² + CD² [ ∵ given AD² = BD.CD ]
= (BD + CD )² = BC²
Thus in triangle ABC we have , AB² + AC²= BC²
hence triangle ABC is a right triangle right angled at A
∠ BAC = 90°
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