b)  \(ad=bc\)

\(\Rightarrow ac-ad+bc-bd=ac-bc+ad-bd\)

\(\Rightarrow a.\left(c-d\right)+b.\left(c-d\right)=c.\left(a-b\right)+d.\left(a-b\right)\)

\(\Rightarrow\left(a+b\right)\left(c-d\right)=\left(c+d\right)\left(a-b\right)\)

\(\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}\)

ai giai gium minh ma cinh xac nhat minh cho

a: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a}{a-b}=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\)

\(\frac{c}{c-d}=\frac{dk}{dk-d}=\frac{dk}{d\left(k-1\right)}=\frac{k}{k-1}\)

Do đó: \(\frac{a}{a-b}=\frac{c}{c-d}\)

c: \(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\)

\(\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\)

Do đó: \(\frac{a}{3a+b}=\frac{c}{3c+d}\)

e: \(\frac{a\cdot b}{c\cdot d}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\)

\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)

Do đó: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)

a) \(VT=\frac{a}{a+c}=\frac{kb}{kb+kd}=\frac{kb}{k\left(b+d\right)}=\frac{b}{b+d}=VP\)

=> đpcm

b) \(VT=\frac{a^2+c^2}{b^2+d^2}=\frac{\left(kb\right)^2+\left(kd\right)^2}{b^2+d^2}=\frac{k^2b^2+k^2d^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)(1)

\(VP=\frac{ac}{bd}=\frac{kb\cdot kd}{bd}=\frac{k^2bd}{bd}=k^2\)(2)

Từ (1) và (2) => VT = VP => đpcm

A B C D E 1 2

a) Vì BC=2 AB

Mà E là trung điểm của BC

=> AB= BE = EC

Xét ΔABD và ΔEBD có:

AB=BE (cmt)

góc A₁ = góc A₂(gt)

BD: cạnh chung

=> ΔABD=ΔEBD (c.g.c)

=> góc ADB= góc EDB

=> DB là tia pg của góc ADE

b) VÌ ΔABD=ΔEBD( cmt)

=> góc BAD= góc BED=90

Mà : góc DEB + góc DEC=180

=> góc DEB= góc DEC

Xét ΔDEB và ΔDEC có:

DE:cạnh chung

góc DEB = góc DEC(cmt)

BE=CE(gt)

=> ΔDEB=ΔDEC(c.g.c)

=> BD=DC

c) Vì ΔDEB=ΔDEC(cmt)

=> góc B₂= góc C

Mà: góc B+ góc C=90

<=> 2 B₂+ góc C=90

<=> 3 góc B₂=90

<=> B₂=30

Vậy: góc C=góc B₂=30; góc B= 2.B₂=2.30=60

thanks bạn nha

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a^2+ac}{c^2-ac}=\frac{\left(bk\right)^2+bk\cdot dk}{\left(dk\right)^2-bk\cdot dk}=\frac{b^2k^2+bd\cdot k^2}{d^2k^2-bd\cdot k^2}=\frac{b\cdot k^2\left(b+d\right)}{d\cdot k^2\left(d-b\right)}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)

\(\frac{b^2+bd}{d^2-bd}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)

Do đó: \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)