In triangle abc, angle abc equal to 90 degree, angle acb equal to 40 degree and BD is perpendicular to AC. Find angle abd, angle cbc and angle Bac
Answers
Step-by-step explanation:
Answer:
Step-by-step explanation:
As per the property of a angle bisector of a triangle, when an angle bisector is drawn in a triangle, the ratio of the opposite sides forming the bisected angle is equal to the ratio of the segments formed by bisector intersecting the opposite side.
In \Delta ACBΔACB
Seg CE bisects \angle ACB
\therefore \frac{AE}{BE}=\frac{AC}{BC}∴
BE
AE
=
BC
AC
Squaring the both side we get
\frac{AE^{2}}{BE^{2}}=\frac{AC^{2}}{BC^{2}}\text{ .................equation-1}
BE
2
AE
2
=
BC
2
AC
2
.................equation-1
In \Delta ABCΔABC
m\angle ACB=90^{\circ}m∠ACB=90
∘
Seg CD hypotenuse AB
As per the theorem on similarity of right angled triangles we know that when a perpendicular is drawn on the hypotenuse the triangle is divided into two parts, As the corresponding angles of this two triangles are equal so we can say that both triangles are similar.
therefore,
\Delta ABC\sim\Delta ADC\sim\Delta BDCΔABC∼ΔADC∼ΔBDC
In \Delta ABCΔABC and \Delta ADCΔADC
\begin{gathered}\frac{AC}{AB}=\frac{AD}{AC}\\\\\Rightarrow AC^{2}=AB\times AD\text{ ..............equation-2}\end{gathered}
AB
AC
=
AC
AD
⇒AC
2
=AB×AD ..............equation-2
In \Delta ABCΔABC and \Delta BDCΔBDC
\begin{gathered}\frac{BC}{AB}=\frac{BD}{BC}\\\\\Rightarrow BC^{2}=BD\times AB\text{ ..............equation-3}\end{gathered}
AB
BC
=
BC
BD
⇒BC
2
=BD×AB ..............equation-3
From equation - 1 we get
\begin{gathered}\frac{AE^{2}}{BE^{2}}=\frac{AC^{2}}{BC^{2}}\\\\\Rightarrow\frac{AE^{2}}{BE^{2}}=\frac{AB\times AD}{BD\times AB}\\\\\Rightarrow\frac{AE^{2}}{BE^{2}}=\frac{AD}{BD}\end{gathered}
BE
2
AE
2
=
BC
2
AC
2
⇒
BE
2
AE
2
=
BD×AB
AB×AD
⇒
BE
2
AE
2
=
BD
AD
\therefore\frac{AD}{BD}=\frac{AE\times AE}{BE\times BE}=\text{ (Proved)}∴
BD
AD
=
BE×BE
AE×AE
= (Proved)
Answer:
Answer:
Given: ABC is a right triangle,
In which \angle ACB = 90^{\circ}∠ACB=90∘
Also, CD \perp ABCD⊥AB and DE \perp ABDE⊥AB
Where D is any point in AB and E is any point in CB.
To prove : CD^2 \times AC= AD \times AB \times DECD2×AC=AD×AB×DE
Proof:
In triangles, ACB and ADC,
\angle CAB\cong \angle DAC∠CAB≅∠DAC ( Reflexive)
\angle ACB\cong \angle ADC∠ACB≅∠ADC ( Right angles)
Thus, By AA similarity postulate,
\triangle ACB\sim \triangle ADC△ACB∼△ADC -------(1)
⇒ \frac{AC}{AD} = \frac{AB}{AC}ADAC=ACAB
⇒ AC^2= AB\times ADAC2=AB×AD -----------(2)
Now, Similarly,
\triangle ACB\sim \triangle CDB△ACB∼△CDB ----------(3)
Now, In triangles CED and CDB,
\angle ECD\cong \angle DCB∠ECD≅∠DCB ( Reflexive)
\angle CED\cong \angle CDB∠CED≅∠CDB ( Right angles)
By AA similarity postulate,
\triangle CED\sim \triangle CDB△CED∼△CDB ----------(4)
By equation (3) and (4),
\triangle ACB\sim \triangle CED△ACB∼△CED -------(5)
Again by equation (1) and (5)
\triangle ADC\sim \triangle CED△ADC∼△CED
\frac{AC}{CD} = \frac{DC}{ED}CDAC=EDDC
CD^2 = AC\times DECD2=AC×DE ------------(6)
Multiplying equation (2) by CD^2CD2 on both sides,
CD^2\times AC^2= CD^2\times AB\times ADCD2×AC2=CD2×AB×AD
From equation (6),
CD^2\times AC^2= AC\times DE \times AB\times ADCD2×AC2=AC×DE×AB×AD
CD^2\times AC= DE \times AB\times ADCD2×AC=DE×AB×AD
CD^2\times AC= AD\times AB\times DECD2×AC=AD×AB×DE
Hence Proved.