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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