In Delta ABC- angle ABC - 90-circ- and P is a point on AC such that angle PBC- - angle PCB- Show that - PA - PB- Enter 1 true else 0

Question

In Delta ABC- angle ABC - 90-circ- and P is a point on AC such that  angle PBC- - angle PCB- Show that - PA - PB- Enter 1 true else 0

Toppr Admin · Accepted Answer

In -x25B3-ABC-x2220-A-x2220-B-x2220-C-180-x2220-A-90-x2220-C-180-x2220-A-90-x2212-x2220-CAlso- -x2220-ABP-x2220-B-x2212-x2220-PBC-x2220-ABP-90-x2212-x2220-PBC-x2220-ABP-90-x2212-x2220-C -Given PB - PC- by Isosceles triangle property- -x2220-PBC-x2220-C-Hence- -x2220-ABP-x2220-Aor- PA-PB

In $Δ A B C, ∠$ $A B C = 90^{\circ}$ and P is a point on AC such that $∠ P B C = ∠ P C B$ . Show that : $P A = P B .$ Enter $1$ if true else $0$

In $△ A B C$ ,
$∠ A + ∠ B + ∠ C = 180$
$∠ A + 90 + ∠ C = 180$
$∠ A = 90 - ∠ C$
Also, $∠ A B P = ∠ B - ∠ P B C$
$∠ A B P = 90 - ∠ P B C$
$∠ A B P = 90 - ∠ C$ (Given PB = PC, by Isosceles triangle property, $∠ P B C = ∠ C$ )
Hence, $∠ A B P = ∠ A$
or, $P A = P B$

In ΔABC, ∠ ABC=90∘ and P is a point on AC such that ∠PBC=∠PCB. Show that : PA=PB. Enter 1 if true else 0

In △ABC,∠A+∠B+∠C=180∠A+90+∠C=180∠A=90−∠CAlso, ∠ABP=∠B−∠PBC∠ABP=90−∠PBC∠ABP=90−∠C (Given PB = PC, by Isosceles triangle property, ∠PBC=∠C)Hence, ∠ABP=∠Aor, PA=PB

In $Δ A B C, ∠$ $A B C = 90^{\circ}$ and P is a point on AC such that $∠ P B C = ∠ P C B$ . Show that : $P A = P B .$ Enter $1$ if true else $0$

In $△ A B C$ ,
$∠ A + ∠ B + ∠ C = 180$
$∠ A + 90 + ∠ C = 180$
$∠ A = 90 - ∠ C$
Also, $∠ A B P = ∠ B - ∠ P B C$
$∠ A B P = 90 - ∠ P B C$
$∠ A B P = 90 - ∠ C$ (Given PB = PC, by Isosceles triangle property, $∠ P B C = ∠ C$ )
Hence, $∠ A B P = ∠ A$
or, $P A = P B$