question_answer

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.