1. \[\Delta ABD\tilde{\ }\Delta CAD\] 2. \[\Delta ABD\cong \Delta CDA\] 3. \[\Delta ADB\tilde{\ }\Delta CAB\] Of these statements the correct ones are combinations of
A) 1 and 2
B) 1 and 3
C) 2 and 3
D) All of these
Correct Answer: D
Solution :
| I. In \[\Delta ABD\] and \[\Delta CAD\] |
| \[\because \] \[\angle ADB=\angle ADC=90{}^\circ \]each |
| \[\angle BAD=\angle ACD\] |
| \[\therefore \] \[\Delta ADB\sim \Delta CAD\] |
| II. In \[\Delta ABD\]and \[\Delta CDA\][\[\because \]each =\[90{}^\circ \]\[-\angle DAC\]] |
| \[\angle ADB=\angle ADC=90{}^\circ \]each |
| \[\angle BAD=\angle ACD=90{}^\circ -\angle DAC\](each) |
| and AD = AD (common) |
| \[\therefore \] \[\Delta ABD\cong \Delta CDA\] |
| III. In \[\Delta ADB\] and \[\Delta CAB\] |
| \[\angle ADB=\angle BAC\,\,90{}^\circ \]each |
| \[\angle B=\angle B\](common) |
| \[\therefore \] \[\Delta ADB\sim \Delta CAB\] |
| Here, I, II and III are correct statements |
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