 1BPD 2(SĐ BD - SĐAC),  1AQC 2(SĐ BD + SĐAC)  BPD AQC  = SĐ BD = 1400  700BCDB) HS TỰ CHỨNG MINH

Question

5. a) Ta có:  1BPD 2(sđ BD - sđAC),  1AQC 2(sđ BD + sđAC)  BPD AQC  = sđ BD = 1400  700BCDb) HS tự chứng minh

Answer

5. a) Ta có:  1BPD 2(sđ BD - sđAC),  1AQC 2(sđ BD + sđAC)  BPD AQC  = sđ BD = 1400  700BCDb) HS tự chứng minh