In the figure if \[\angle ABD=\angle XYD=\angle CDB=90{}^\circ .\text{ }AB=a,XY=c\] and \[CD=b\], then prove that\[c\text{ (}a+b)=ab\]. |
Answer:
To prove: \[c\text{ (}a+b)=ab\] In \[\Delta \text{ }ABD\]& \[\Delta \text{ }DXY\] \[\angle B=\angle XYD\] [Each \[90{}^\circ \]] \[\angle XDY=\angle ADB\] [Common] So by AA similarity \[\Delta \text{ }DAB\tilde{\ }\Delta \text{ }DXY\] \[\therefore \] \[\frac{DY}{DB}=\frac{XY}{AB}\] \[DY=\frac{c}{a}(BD)\] ?(i) In \[\Delta \text{ }BCD\] & \[\Delta \text{ }BYX\] \[\angle XYB=\angle D\] [Each \[90{}^\circ \]] \[\angle CBD=\angle XBY\] [Common] So by AA similarity, \[\Delta \text{ }BYX\tilde{\ }\Delta \text{ }BDC\] \[\frac{BY}{BD}=\frac{XY}{CD}\] \[BY=\frac{c}{b}(BD)\] ...(ii) Adding eq. (i) and eq. (ii) \[DY+BY=-\,(BD)+\frac{c}{b}(BD)\] \[BD=BD\left[ \frac{c}{a}+\frac{c}{b} \right]\] \[\frac{BD}{BD}=\left[ \frac{cb\,\,\,\,\,ca}{ab} \right]\] \[1=\frac{c(a+b)}{ab}\] \[c(a+b)=ab\] Hence Proved
You need to login to perform this action.
You will be redirected in
3 sec