In any triangle ABC ,the bisector of angle A cuts BC in D show that: (AD)^2=AB*AC - BD*CD?

Answer 1

Draw the triangle with A at the top, the base along the bottom with B at the left and C at the right. Let x be the angle in the triangle on the left where the bisector of A cuts BC. So the similar angle in the triangle on the right will be 180 - x.

Use the law of cosines on the two triangles.

Left Triangle: AB^2 = BD^2 + AD^2 - 2(BD)(AD)cos(x)

Right Triangle: AC^2 = DC^2 + AD^2 - 2(DC)(AD)cos(180-x)
This is: AC^2 = DC^2 + AD^2 + 2(DC)(AD)cos(x)

Solve each for cos(x) and set these equal to each other.

(AB^2 - BD^2 - AD^2)/(-2(BD)(AD)) = (AC^2 - DC^2 - AD^2 )/(2(DC)(AD))
AD^2(BD + DC) = BD(AC^2 - DC^2 ) + DC(AB^2 - BD^2)
AD^2(BD + DC) = BD(BD*DC + AC^2) - DC(BD*DC + AB^2)

AD^2(BD + DC) = BD*AC^2 + DC*AB^2 - BD*DC(BD + DC) ....... [Eqn 1]

Now use law of sines twice for each triangle. If a is the angle at vertex A then:
For the left triangle:
sin(a)/BD = sin(x)/AB

For the right triangle:
sin(a)/DC = sin(180-x)/AC = sin(x)/AC

Combine these by using sin(a)/sin(x) and get:
BD/AB = DC/AC

Use BD= AB*DC/AC and BD*AC^2 from [Eqn 1] becomes:
AC*AB*DC

Use AB = BD*AC/DC and DC*AB^2 from [Eqn 1] becomes:
AC*AB*BD

So (BD*AC^2 + DC*AB^2) becomes AC*AB(BD + DC) and:
AD^2(BD + DC) = AC*AB(BD + DC) - BD*DC(BD + DC)

AD^2 = AC*AB - BD*DC

Answer 2

no longer that no longer ordinary. Label attitude BAC as a, ABC as b, and so on. look at triangle ABD. It has angles a/2, b, and d, the place a/2 + b + d = one hundred eighty. (d = attitude ADB) additionally, a + b + c = one hundred eighty, so placing those equivalent we get a/2 + b + d = a + b + c d = a/2 + c. Now the attitude opposite area BD in ABD is a/2, and opposite AB is d = a/2 + c. because c is > 0 this suggests d > a/2, so by way of the regulation of sines BD < AB ------------- sshanbhag, that may no longer acceptable, that's honestly conceivable for BD > AB whether that isn't any longer bisected. think of a triangle with an exceedingly short AB and an exceedingly long BC. for many D between B and C BD would be longer than AB.

Answer 3

Well done, Captain Mephisto!
I have another solution, not involving trigonometry.

Draw a line from D meeting AC at E, such that
angle ADE = angle ABD.

Then triangles ABD, ADE are similar (equiangular because
angle BAD = angle DAE
and ADE = ABD)

Also, since angle ADC = angle ABD + angle BAD (exterior angle of triangle ABD equals sum of interior opposite angles),
therefore angle EDC = angle BAD = angle DAC

Now triangles ADC, DEC have angles DAC, EDC equal, and have angle C in common (i.e. DCA = ECD), and so they are similar.

Now in the similar triangles ABD, ADE,
AD/AE = AB/AD
and so
AD^2 = AB*AE
..........= AB*(AC - EC)
..........= AB*AC - AB*EC

Now in the similar triangles DAC. EDC.
EC/DC = ED/AD, and so we replace EC by
DC*ED/AD.
Hence
AD^2 = AB*AC - AB*DC*ED/AD
But AB/AD = BD/DE (similar triangles ABD, ADE)
and so
AD^2 = AB*AC - BD*DC*ED/DE
...........= AB*AC - BD*DC