In \[\Delta \,ABC\], altitudes AD and CE intersect each other at the point P. Prove that |
(i) \[\Delta \,APE\sim \Delta \,CPD\] |
(ii) \[AP\times PD=CP\times PE\] |
(iii) \[\Delta \,ADB\tilde{\ }\Delta \,CEB\] |
(iv) \[AB\times CE=BC\times AD\] |
Answer:
Given, In \[\Delta \,ABC,\text{ }AD\bot BC\]and \[CE\bot AB\] (i) In \[\Delta \,APE\] and \[\Delta \,CPD\] \[\angle \text{1}=\angle 4\] [Each \[90{}^\circ \]] \[\angle 2=\angle 3\] [Vertically opposite angles! By AA axiom \[\Delta \,APE\tilde{\ }\Delta \,CPD\] Hence Proved. (ii) \[\Delta \,APE\tilde{\ }\Delta \,CPD\] [Proved above] \[\therefore \frac{AP}{CP}=\frac{PE}{PD}\] [CPCT] \[\Rightarrow \,AB\times PD=CP\times PE\] Hence Proved (iii) In \[\Delta \,ADB\sim \Delta \,CEB\] \[\therefore \frac{AB}{CB}=\frac{AD}{CE}\] [cpct] \[AB\times CE=BC\times AD\] Hence Proved. \[={{(\cot \,A+\sec \,B)}^{2}}-{{(\tan \,B-\cos ec\,A)}^{2}}\] \[\angle 5=\angle 7\] (Each \[90{}^\circ \]) \[\angle 6=\angle 6\] (Common) By AA axiom, \[\Delta \,ADB\tilde{\ }\Delta \,CEB\] Hence Proved. (iv) \[\Delta \,ADB\tilde{\ }\Delta \,CEB\] [Proved Above] \[\frac{AB}{CB}=\frac{AD}{CE}\] [cpct] \[\Rightarrow AB\times CE=BC\times AD\] Hence Proved.
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