ABC is a right angled triangle which is right angeled at c AB=c BC =aAB =c and p id perpendicular from C on AB PT c=ab÷p
Answers
Step-by-step explanation:
Here, ABC is a right triangle,
In which ∠C = 90°, AB =c, AC= b, BC = a,
Also, p is the length of the perpendicular from c to ab,
We have to prove that:
\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}p21=a21+b21
Proof:
Let D be the point on the segment AB,
Such that CD = p,
In triangle ACB and ADC,
\angle CAB\cong \angle DAC∠CAB≅∠DAC ( reflexive )
\angle ACB\cong \angle ADC∠ACB≅∠ADC ( Right angle )
By AA similarity postulate,
\triangle ACB\sim \triangle ADC△ACB∼△ADC
By the property of similar triangles,
\frac{AC}{CB}=\frac{AD}{DC}CBAC=DCAD
\frac{b}{a}=\frac{AD}{p}ab=pAD
\frac{pb}{a}=ADapb=AD ------------ (1)
Now, by the Pythagoras theorem,
AC^2=AD^2+CD^2AC2=AD2+CD2
b^2=(\frac{pb}{a})^2+p^2b2=(apb)2+p2
b^2=\frac{p^2b^2}{a^2}+p^2b2=a2p2b2+p2
1=\frac{p^2}{a^2}+\frac{p^2}{b^2}1=a2p2+b2p2
\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}p21=a21+b21
Hence, proved.
