Please solve the following and give a step-by- step explanation.
Answers
Answer:
We have,
ABC is a right angle triangle.
In which \angle ACB={{90}^{o}}∠ACB=90
o
And CD\bot AB\,and\,DE\bot ABCD⊥ABandDE⊥AB
Where D is any point in AB and E is any point in CB.
Prove that:- C{{D}^{2}}\times AC=AD\times AB\times DECD
2
×AC=AD×AB×DE
Proof:-
In\,\Delta ACB\,and\,\Delta ADCInΔACBandΔADC
\angle CAB\cong \angle DAC\,\,\left( \text{Reflexive} \right)∠CAB≅∠DAC(Reflexive)
\angle ACB\cong \angle ADC\,\left( \text{Right}\,\text{angle} \right)∠ACB≅∠ADC(Rightangle)
Thus, By AA similarity.
Then,
\Delta ACB\sim \Delta ADC\,\,\,......\,\,\left( 1 \right)ΔACB∼ΔADC......(1)
\Rightarrow \dfrac{AC}{AD}=\dfrac{AB}{AC}⇒
AD
AC
=
AC
AB
\Rightarrow A{{C}^{2}}=AB\times AD\,......\,\,\left( 2 \right)⇒AC
2
=AB×AD......(2)
Now, Similarly
\Delta ACB\sim \Delta CDB\,\,......\,\,\left( 3 \right)ΔACB∼ΔCDB......(3)
Now,
In\,\Delta CED\,and\,\Delta CDBInΔCEDandΔCDB
\angle ECD\cong \angle DCB\,\,\left( \text{Reflexive} \right)∠ECD≅∠DCB(Reflexive)
\angle CED\cong \angle CDB\,\left( \text{Right}\,\text{angle} \right)∠CED≅∠CDB(Rightangle)
By AA similarity.
\Delta CED\sim \Delta CDB\,\,.......\,\,\left( 4 \right)ΔCED∼ΔCDB.......(4)
From equation (3) and (4) to,
\Delta ACB\sim \Delta CED\,\,......\,\,\left( 5 \right)ΔACB∼ΔCED......(5)
By equation (1) and (5) to, we get,
\Delta ADC\sim \Delta CEDΔADC∼ΔCED
\dfrac{AC}{CD}=\dfrac{DC}{ED}
CD
AC
=
ED
DC
C{{D}^{2}}=AC\times DE\,\,......\,\,\left( 6 \right)CD
2
=AC×DE......(6)
From equation (2) to,
A{{C}^{2}}=AB\times ADAC
2
=AB×AD
On multiplying both side by C{{D}^{2}}CD
2
and we get,
C{{D}^{2}}\times A{{C}^{2}}=C{{D}^{2}}\times AB\times ADCD
2
×AC
2
=CD
2
×AB×AD
C{{D}^{2}}\times A{{C}^{2}}=AC\times DE\times AB\times ADCD
2
×AC
2
=AC×DE×AB×AD
C{{D}^{2}}\times AC=DE\times AB\times ADCD
2
×AC=DE×AB×AD
C{{D}^{2}}\times AC=AD\times AB\times DECD
2
×AC=AD×AB×DE
Hence proved.