Question

show that AB2 = AD.AC

Answer 1

Solution :-

In ∆ADB and ∆ABC we have,

→ ∠ADB = ∠ABC { Each 90° }

→ ∠BAD = ∠CAB { Common. }

So,

→ ∆ADB ~ ∆ABC { By AA similarity . }

then,

→ AD/AB = AB/AC { when two ∆'s are similar there corresponding sides are in same proportion. }

therefore,

→ AB * AB = AD * AC

→ AB² = AD•AC (Proved.)

Extra knowledge :-

To Prove :- BC² = CD•CA

In ∆BDC and ∆ABC we have,

→ ∠BDC = ∠ABC { Each 90° }

→ ∠BCD = ∠ACB { Common. }

So,

→ ∆BDC ~ ∆ABC { By AA similarity . }

then,

→ DC/BC = BC/AC { when two ∆'s are similar there corresponding sides are in same proportion. }

therefore,

→ BC * BC = DC * AC

→ BC² = DC•AC

→ BC² = CD•CA (Proved.)

Note :- Now, try to prove BD² = AD•DC also by making ∆ADB and ∆CDB similar .

Learn more :-

in triangle ABC seg DE parallel side BC. If 2 area of triangle ADE = area of quadrilateral DBCE find AB : AD show that B...

https://brainly.in/question/15942930

2) In ∆ABC seg MN || side AC, seg MN divides ∆ABC into two parts of equal area. Determine the value of AM / AB

https://brainly.in/question/37634605

Answer 2

Step-by-step explanation:

Given:Figure

To find: Show that AB²=AD.AC

Solution:

In ∆ABC and ∆ABD

$\angle ABC =\angle ADB$

$\angle BAC=\angle BAD$ common in both.

Thus,

By AA CRITERION of similarity both triangles CBA and BDA are similar.

We know that if two triangles are similar than ratio of their corresponding sides are equal.

$\frac{BC}{BD}=\frac{AB}{AD}=\frac{AC}{AB}$

From last two

$\frac{AB}{AD}=\frac{AC}{AB}$

cross multiply

AB.AB= AD.AC

AB²=AD.AC

Final answer:

AB²=AD.AC

has been shown.

Hope it helps you.

To learn more on brainly:

Plot the point P(– 6, 2) and from it draw PM and PN as perpendicular to x-axis and y-axis respectively. Write coordinate...

https://brainly.in/question/5909135

A traffic police signal board is in the shape of a triangle marked with "SCHOOL AHEAD" displayed on the road side. The v...

https://brainly.in/question/47222204