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In ΔABC, if ∠B = 90°, BD ⊥ AC, then 

(A) ΔADB ~ ΔBDC 

(B) ΔADB ~ ΔABC 

(C) ΔBDC ~ ΔABC 

(D) All of these

2 Answers

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Correct option is (D) All of these

Given that \(\angle B=90^\circ\) & \(BD\bot AC\)

\(\therefore\) \(\angle A+\angle B+\angle C=180^\circ\)   (In \(\triangle ABC)\)

\(\Rightarrow\) \(\angle A+\angle C=90^\circ\)  _______________(1)   \((\because\angle B=90^\circ)\)

Let \(\angle ABD=\angle 1\;\&\;\angle CBD=\angle 2\) 

In \(\triangle ADB,\)

\(\angle A+\angle ADB+\angle 1=180^\circ\)

\(\Rightarrow\) \(\angle A+\angle 1=90^\circ\) _______________(2)   \((\because\angle ADB=90^\circ\,as\,BD\bot AC)\)

From (1) & (2), we obtain

\(\angle C=\angle1\)   _______________(3)

(A) In right \(\triangle ADB\;\&\;\triangle BDC,\)

\(\angle C=\angle1\)                                      (From (3))

\(\angle ADB=\angle BDC=90^\circ\)               \((BD\bot AC)\)

\(\therefore\) \(\triangle ADB\sim\triangle BDC\)    (By AA similarity rule)

(B) In right \(\triangle ADB\;\&\;\triangle ABC,\)

\(\angle DAB=\angle BAC\)               (Common angle)

\(\angle ADB=\angle ABC=90^\circ\)                    (Given)

\(\therefore\) \(\triangle ADB\sim\triangle ABC\)    (By AA similarity rule)

(C) \(\because\) \(\triangle ADB\sim\triangle BDC\) & \(\triangle ADB\sim\triangle ABC\)

\(\therefore\) \(\triangle BDC\sim\triangle ABC\)

+1 vote
by (35.1k points)

Correct option is: (D) All of these

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